If the foot of the perpendicular from $(0,0,0)$ to a plane is $(1,2,3)$, then the equation of the plane is

The vector equation of the plane which is at a distance of $ \frac{3}{\sqrt{14}} $ from the origin and the normal vector from the origin is $ 2 \hat{i}-3 \hat{j}+\hat{k} $ is:

$A$ tetrahedron has vertices $O(0,0,0)$, $A(1,2,1)$, $B(2,1,3)$, and $C(-1,1,2)$. If $\theta$ is the angle between the faces $OAB$ and $ABC$, then $\cos \theta =$

If $\overrightarrow{p} = 4\hat{i} - \hat{j} + \hat{k}$ is a point and $\overrightarrow{q} = 9\hat{i} - 2\hat{j} + 6\hat{k}$ is a normal vector,then the perpendicular distance of the origin from the plane passing through $\overrightarrow{p}$ and perpendicular to $\overrightarrow{q}$ is

If $M$ is the foot of the perpendicular drawn from $P(1,2,-1)$ to the plane passing through the point $A(3,-2,1)$ and perpendicular to the vector $4 \hat{i}+7 \hat{j}-4 \hat{k}$, then the length of $PM$ is

The coordinates of the foot of the perpendicular drawn from the origin to the plane $3x + 2y + 6z = 56$ are: